The truss is modeled in 2D space (x and y axes). For 2D systems, the stiffness matrix of a truss member is a 4x4 matrix. The coefficients of this matrix depend on the orientation of the member, specifically the sine and cosine of the angle it makes. When a truss member is horizontal or vertical, some stiffness coefficients become zero due to the sine or cosine of the angle being zero. This holds true even though we are working with a 2D model. For truss analysis, we don’t have 1D models; the models are either 2D (plane truss) or 3D (space truss).

@DrStructure Thank you very much for the kind answer :) 💖

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Yes, in the free online course referenced in the video description field.

In the loading case discussed in the lecture, no bending moment develops in the frame. However, consider a vertical distributed load applied to the horizontal member of the frame. This type of loading generates a bending moment, much like a distributed load induces bending moment in a simply supported beam.

We are treating the hangar as a closed structure, with the hangar door kept closed during high winds. Consequently, the wind impact on the hangar is identical for both the North-South and South-North directions.

10 is the point load magnitude. The load produces a moment about the cut point. The moment arm is (x-4) where x is the distance from the left end of the beam to the cut point and 4 is the distance from the left end of the beam to the point of application of the point load. So, the distance from the point of application of the point load to the cut point (the moment arm) is x-4.

Correct. In their current state, ChatGPT and other generative AI tools are not capable of effectively solving complex physics or mechanics problems. They often make mistakes and generally lack the capability to handle such problems with sufficient accuracy. That said, there is potential to customize these tools to function as effective tutoring systems, which is the long-term goal of our project.

The reactions at A and F are correct. The reaction at D is 3 x 5 = 15. The influence line for the reaction force at D forms a triangle with a height of 1 at D and a height of 3 at E. When D is pushed up by a unit, E rises correspondingly by three units. As a result, segment CD rotates counterclockwise (about C) , while segment EF rotates clockwise (about F).

@DrStructure For the reaction at D, I try to figure out how at E the height is 3 if at D it's only 1. Isn't it height at E is just 3/7? Thus, the reaction at D = 1x5 = 5kN when load is at D and it's 3/7x5 = 15/7kN when load is at E.

When point D is displaced upward by 1 unit, point C remains stationary but rotates due to the movement. This rotation causes point E to move upward. A right triangle forms with the base between points C and E, which is 6 meters, and the height at E, which we’ll refer to as h. The distance between points C and D is 2 meters. From the geometry of the triangle, the relationship between the displacement at D (1 unit) and the base and height of the triangle is given by: 1/2 = h/6. Solving for h: h = (1/2) * 6 = 3. The upward movement of point E results in a support reaction at D. If the load applied at E is 5 (units of force), the reaction at D can be calculated as: Reaction at D = h * Load at E = 3 * 5 = 15.

In structural engineering, the terms second moment of area and moment of inertia are often used interchangeably because, for most structural applications, the mass distribution across the cross section remains constant. This consistency allows us to focus solely on the geometry of the cross-section rather than on varying mass, which simplifies the analysis. The second moment of area, which is a measure of a cross-section’s resistance to bending, depends only on the shape and dimensions of the cross-section. Since the material density doesn’t typically change across the cross-section of structural elements, the second moment of area becomes the primary factor affecting bending and deflection. In this context, “moment of inertia” is commonly used as a shorthand, even though it technically refers to mass moment of inertia in physics. Because structural engineers rarely need to account for changing mass within the cross-section, they often use the term “moment of inertia” to describe the second moment of area. This interchangeable use has become common practice, as it reflects the fact that structural elements are evaluated mainly on their geometric properties and their resistance to bending, not on rotational dynamics where mass distribution would be crucial.

@@DrStructure Thanks!

The stiffness matrix shown at 1:50 is in the global coordinate system, meaning the transformation matrix is embedded within it. D1 represents the force in the x direction, not in the local (member) coordinate system, but in the global coordinate system. Therefore, the solution (displacements and member forces) is entirely in the global coordinate system.

@ Thank you for clarification and great video lectures!

The aim of this lecture is to illustrate the process of constructing and using free-body diagrams. We do not have a lecture that covers a complete analysis of the system. It may be worthwhile to prepare such a lecture as a continuation of this presentation.

ThetaA, ThetaB, and ThetaC are the unknowns in the three equilibrium equations. Thus, we have three equations with three unknowns, which can be solved using linear algebra. For example, a method like Gaussian Elimination can be used to find the values of these unknowns.

To draw the moment influence line for the fixed support at A, we introduce a fictitious hinge at A, allowing rotation at this support. At point C, adding a fictitious hinge enables rotation of segments BC and CD, while segments AB and BF remain stationary. For point E, placing a fictitious hinge allows rotation of segments BD and BE. Here is the solution video for Problem 1: kzbin.info/www/bejne/eaLEoIqIa7FlqsU

@DrStructure Thanks a lot. Therefore, as I understand now, fictitious hinge at . . . 1) fixed support allows rotation, 2) roller or hinge support allows rotation but not vertical movement?

Correct!

Yes, the bending moment at a hinge is always zero, regardless of the beam’s boundary conditions.

@@DrStructure Thank you for the reply. This is just a follow up question to illustrate that I got the information correctly. A beam shown at 6:20 mark, with two segments i.e. AC and CD. If for example, I would like to find out the moment at the real internal hinge at point C is zero indeed. Segment CD will only move upward if beam AC allows it. Since points A and B won't, therefore the moment at point C (the real hinge) is zero indeed. Is it correct?

Yes, that is correct. We can draw the influence line for the moment at point C (the location of the real hinge) to confirm that the moment at the hinge remains zero, regardless of the load’s position. Since there is a real hinge at C, there’s no need to place a fictitious hinge there. To draw the moment influence line at C, we apply the pair of moments at that point and depict the displaced shape of the beam. Joint C cannot move vertically, as segment AC, supported by a pin and roller at points A and B, remains unable to rotate. Therefore, the influence line is a straight line at a height of zero, indicating that the moment at C stays zero as the unit load travels across the beam.

Please be more specific. Do you mean how we draw the influence line for a diagonal member of the truss? Similar to the example (member BC) presented in the lecture.

When considering determinacy, a beam resting on a pin and two rollers is typically considered indeterminate. Why? Because it has four support reactions but there are only three static equilibrium equations. So, why is it that, in this case, we can determine the support reactions? What condition allows us to simplify the problem, as you’ve suggested? How would you approach the problem if that condition were removed?

​@@DrStructure I think the fact that it is overhang simplifies the problem, as there is no reaction/internal moment or shear force at D, therefore for segment CD we have 2 equations and 2 unknowns to solve for internal shear force and moment. It's like hinged beam, where globally indeterminate structure can be solved in parts by statics. Is that what you mean?

Not quite. It is true that we can simplify the problem by replacing the overhang with a force and a moment applied at C. So, we know the moment at C is wL^2/2. This is the moment due to the load applied on the overhang. But that does not enable us to determine the reaction forces at A, B, and C. There are still too many unknowns to be solved using the static equilibrium equations. Let’s simplify the problem by removing the overhang altogether. Now, we have a two-span beam with a pin support at A, and rollers at B and C, resulting in four unknown reaction forces: Ax, Ay, By, and Cy. Can we determine these unknowns using the static equilibrium equations? We know Ax=0. But what about Ay, By, and Cy? How do we determine them?

Thank you for your thoughtful comment! 1. You’re absolutely correct that elastomeric bearings in practical applications don’t exhibit such large deflections. In this video, we intentionally used a case with exaggerated deflection to clearly demonstrate the impact of support settlement on reaction forces and to illustrate how we can model supports as springs. This approach is designed to enhance understanding of the analysis process, rather than to replicate real-world conditions. 2. As for your point about the number of unknowns, we completely agree that handling more than 2 unknowns by hand can become challenging, regardless of the analysis method. In practice, we typically rely on structural analysis software to streamline these calculations. However, it’s beneficial from a learning perspective to go through the process manually, as it deepens understanding. For those interested in exploring more complex cases without structural analysis software, tools like MATLAB or Mathematica are great for quickly solving systems of equations, even when we model and solve problems by hand. Thanks again for your insights-they add valuable context to the presentation.

@@DrStructure Many many years ago I wrote a computer program to do structural analysis by general matrix methods. My first frame analysis test problem was giving really strange results. I finally discovered that my input data was the problem- I had made up some numbers for material properties and the interior columns were too small allowing a lot of deflection and changing the bending moments. It was like an exaggerated version of the problem in your video.

The solution for the exercise problem is provided in the free online course referenced in the video description field.

Correct, we need to calculate the deflection when the a vertical reaction is the redundant forces.

Yes, if you don’t have access to your notes or a textbook, to apply the method, you need to remember the two slope-deflection equations as well as the equations for calculating the fixed-end moments.

@@DrStructure is it practical? I mean anyone could do it if he knows the formula unlike the force method

Yes, it is practical. When you use the method several times, the equations can be recalled from memory with ease.

You are correct. We have updated the solution file. Thank you for bringing this error to our attention.

Please elaborate and justify your assertion. The pressure coefficients (GCp) have been determined through wind tunnel experiments as a function of the area of the model canopy used in the testing. The wind provisions in the ASCE manual provide wind pressure values for surfaces subjected to wind, whether they are roofs, walls, or canopies. These pressure values represent the force per unit area on the system. It's important to note that the method by which this pressure is transmitted to the supporting structure is not a factor in determining the wind pressure on the system itself.

To solve for the unknowns, including Ey, we need to apply linear algebra techniques. Specifically, we can use established methods such as Gaussian Elimination to solve the system of linear equations.

The solutions for the exercise problems are available in the free online course referenced in the video description field.